Question

\(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\)

Answer 1

\(x\ne\left\{-2;\pm5\right\}\)

\(\frac{x+9}{\left(x+2\right)\left(x-5\right)}-\frac{x+15}{\left(x+5\right)\left(x-5\right)}-\frac{1}{x+2}=0\)

\(\Leftrightarrow\frac{\left(x+9\right)\left(x+5\right)}{\left(x+2\right)\left(x^2-25\right)}-\frac{\left(x+15\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-25\right)}-\frac{x^2-25}{\left(x+2\right)\left(x^2-25\right)}=0\)

\(\Leftrightarrow x^2+14x+45-\left(x^2+17x+30\right)-x^2+25=0\)

\(\Leftrightarrow-x^2-3x+40=0\Rightarrow\left[{}\begin{matrix}x=5\left(l\right)\\x=-8\end{matrix}\right.\)